I'm currently enrolled in a introductory course on logic and I'm until now everything has been going great. I'm having some trouble with applying monotonicity and the strengthening/weakening of propositions.

The general of concept of strengthening/weakening seems clear; I understand is a stronger proposition than , so: , since the former has 'less 1s' in its truth table it's a stronger/more restrictive proposition.

Monotonicity is a bit foggier. How I see it, is that it's basically a form of substitution if we have , then . If we substitute the left hand side with the 'same thing' we substitute the right hand side with, the relation with respect to stronger/weaker holds. Any more clarification on this subject is more than welcome.

So much for theory. Now I'm asked to show via a calculation that is a tautology. To do this I have to show that .

Using some basic calculation/rewriting rules for implication, de Morgan and double negation, I ended up here:

Now it gets messy, but by applying distribution to the left hand side twice I end up with this:

Since the basic rules of strengthening\weakening tell us that , we also have that .

By applying weakening to: , we have that: , which is what we wanted.

Now, for me this makes sense, but is this actually correct? (Giggle)

Edit: I'd like to add another exercise I'm not quite sure of, since it seems to be a bit harder.

Show that: is a tautology. Again, to show this the left hand side should equal the right hand side.

First step is rewriting the implications:

By weakening we now have:

But I'm kind of stuck here.
I could distribute and apply strengthening, but I'm not sure that would be legal. Any help would be welcome!

October 8th 2012, 06:18 AM

emakarov

Re: Showing a formula is a tautology

Quote:

Originally Posted by Diligo

So much for theory. Now I'm asked to show via a calculation that is a tautology. To do this I have to show that .

The implication is a tautology, but the biconditional is not: consider R = False and Q = True.

October 8th 2012, 06:40 AM

Diligo

Re: Showing a formula is a tautology

I've made an error while typing this, wherever you see it should read . But I'm unable to edit my post :(

October 8th 2012, 01:33 PM

emakarov

Re: Showing a formula is a tautology

In calculus, a function f is called monotonic if x <= y implies f(x) <= f(y) for all x, y. Monotonicity in Boolean logic is the same as monotonicity of functions when Boolean formulas are considered as expressions defining function from {0, 1} to {0, 1}, where 0 represents False and 1 represents True. For example, P /\ R defines the function min(P, R). If we keep R constant, this function is monotonic in P, i.e., if P <= Q, then min(P, R) <= min(Q, R).

I believe your solution to the first problem makes sense. However, your method is a curious mix of using equivalences and logical implications. Why do you convert some => into disjunctions but other => into |=? Are you required to use this method?

I personally try not to convert P => Q into ~P \/ Q. Here is how to show R /\ (P => Q) => (R => P) => Q treating implication as "if ... then ...". Assume (1) R, (2) P => Q and (3) R => P. From (3) and (1) we get P, and from it and (2) we get Q, as required. The second formula can be proved in a similar way.

October 10th 2012, 05:19 AM

Diligo

Re: Showing a formula is a tautology

Well basically in this part of the course we are required to use these steps, the second part is more about reasoning, more in line with your method :)